Solve (12-5)^2+(7-6)^2=r^2

$$( 12 - 5 ) ^ { 2 } + ( 7 - 6 ) ^ { 2 } = r ^ { 2 }$$

Solve for r

$$r=5\sqrt{2}\approx 7.071067812$$

$$r=-5\sqrt{2}\approx -7.071067812$$

Steps by Finding Square Root

Subtract 5 from 12 to get 7.

$$7^{2}+\left(7-6\right)^{2}=r^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49+\left(7-6\right)^{2}=r^{2}$$

Subtract 6 from 7 to get 1.

$$49+1^{2}=r^{2}$$

Calculate 1 to the power of 2 and get 1.

$$49+1=r^{2}$$

Add 49 and 1 to get 50.

$$50=r^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$r^{2}=50$$

Take the square root of both sides of the equation.

$$r=5\sqrt{2} r=-5\sqrt{2}$$

Steps Using the Quadratic Formula

Subtract 5 from 12 to get 7.

$$7^{2}+\left(7-6\right)^{2}=r^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49+\left(7-6\right)^{2}=r^{2}$$

Subtract 6 from 7 to get 1.

$$49+1^{2}=r^{2}$$

Calculate 1 to the power of 2 and get 1.

$$49+1=r^{2}$$

Add 49 and 1 to get 50.

$$50=r^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$r^{2}=50$$

Subtract 50 from both sides.

$$r^{2}-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{0±\sqrt{0^{2}-4\left(-50\right)}}{2}$$

Square 0.

$$r=\frac{0±\sqrt{-4\left(-50\right)}}{2}$$

Multiply -4 times -50.

$$r=\frac{0±\sqrt{200}}{2}$$

Take the square root of 200.

$$r=\frac{0±10\sqrt{2}}{2}$$

Now solve the equation r=\frac{0±10\sqrt{2}}{2} when ± is plus.

$$r=5\sqrt{2}$$

Now solve the equation r=\frac{0±10\sqrt{2}}{2} when ± is minus.

$$r=-5\sqrt{2}$$

The equation is now solved.

$$r=5\sqrt{2} r=-5\sqrt{2}$$